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Math Help - use reduction of order to solve the problem...

  1. #1
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    use reduction of order to solve the problem...

    y''-4y=2; y1=e^(-2x)

    i dont see how my text got yp=1/2

    ok so this is what i have so far....

    let y2=ue^(-2x)





    so









    so



    i have the yc part i just dont know how to get the yp..what am i missing or not getting??
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  2. #2
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    x''(t)-4x(t)=2. First find the homogeneous x''(t)-4x(t)=0, characteristic equation m^2-4=0, m1/2=+-2, xh=C1*e^(m1*t)+C2*e^(m2*t)=C1e^(2*t)+C2*e^(-2*t). As we see that a non homogeneous part in start equation is 2, xp must be constant. x=xh+xp. Substitute this in start equation (C1*e^2t+C2*e(-2t)+xp)''-4(C1*e^2t+C2*e(-2t)+xp)=2, xp''=0, -4xp=2, xp=-1/2.


    Kind regards,
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  3. #3
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    derdack, yes, the problem can be solved that way, but I don't believe it is a god idea to advise a person to ignore the teacher's or text's instructions. Slapmaxwell1 says he was instructed to use "reduction of order". It is not at all uncommon to ask a person to solve a simple problem using a more complicated technique in order to practice that technique.

    Slapmaxwell1, yes, letting y= ue^{2x} reduces the left side of the equation to e^{-2x}(u''- 4u') but you appear to have left out the right side of the equation. You have e^{-2x}(u''- 4u')= 2 so that u''- 4u'= 2e^{-2x} and, letting w= u', w'- 4w= 2e^{-2x}.

    That is now a first order equation linear equation. It has e^{-4x} as integrating factor:
    e^{-4x}w'- 4e^{-4x}w= (e^{-4x}w)'= 2e^{-4x}
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  4. #4
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    You are right, but it is completly corect that we have other ways for solving which is important for learner. (John - Beautiful mind) .
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  5. #5
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    Quote Originally Posted by derdack View Post
    You are right, but it is completly corect that we have other ways for solving which is important for learner. (John - Beautiful mind) .
    True, but I was taught reduction of order before your method. The different methods come in their own itme.

    -Dan
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  6. #6
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    Yes. Both of us are right...
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  7. #7
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    i am starting to see why you guys were so pissed that i wasnt using latex...I apologize. learning the commands in beginning is a bit troublesome and take a lil extra time, but its totally worth the effort.
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  8. #8
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    ok ok i see it now.

    if i let y_{p}=A then y'=0 , y''=0

    ok so 0-4A=2 so A=\frac{-1}{2}

    there for y_{p}=\frac{-1}{2} so y=y_{c}+y_{p} because i found Yc I would just put it all together for my final answer. thanks again!
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